Parametric equations of a line

Question

"Find the parametric equations of a line that passes through point $(1,1,0)$, parallel to plane $2x+3y+z=7$ and perpendicular to the line $\frac{x-1}{-2}= \frac{y}{3}=-z-2$"

I don't know where to start, I know how to find the equations when given two different points in space.

score 1 · Accepted Answer · answered Jul 01 '15 at 01:12

1

The equation of the line is $x=1+at$, $y=1+bt$, $z=ct$. The vector $\vec{d}=(a,b,c)$ is parallel to the line, so it must be orthogonal to $\vec{n}=(2,3,1)$. So $2a+3b+c=0$. A parallel vector to the line $\frac{x-1}{-2}= \frac{y}{3}=-z-2$ is $(-2,3,-1)$, so $(a,b,c)\cdot (-2,3,-1)=0$. Hence $-2a+3b-c=0$. So combining this with $2a+3b+c=0$, and adding , we get $6b=0$ so $b=0$.We also get $2a=c$. We can let $c=1$ and $a=2$. and $x=1+2t,y=1,z=t$. I hope I made no mistake!

answered Jul 01 '15 at 01:12

mich95

8,713

I understand why $d$ would be parallel to the line, but where does $n$ come from? The plane is defined by $2x+3y+z=7$, and I'm assuming that's the reason behind the value of $n$. – Nicolás Siplis Jul 01 '15 at 01:41
$\vec{n}$ is a normal vector to the plane. – mich95 Jul 01 '15 at 01:43
Ohhhh, completely forgot about that, excellent! Thank you very much :) – Nicolás Siplis Jul 01 '15 at 01:45

Parametric equations of a line

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